Question

A parallel plate capacitor having capacitance $12pF$ is charged by a battery to a potential difference of $10V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates the work done by the capacitor on the slab is :

• A. 692 pJ
• B. 60 pJ
• C. 508 pJ
• D. 560 pJ

Answer 1

Answer: (C)

Intial energy of capacitor
$U_i=\frac{1}{2}\frac{v^2}{c}$
$=\frac{1}{2}\times \frac{120\times 120}{12}=600J$
Since battery is disconnected so charge remain same.
Final energy of capacitor
$U_f=\frac{1}{2}\frac{v^2}{c}$
$=\frac{1}{2}\times \frac{120\times 120}{12\times 6.5}=92$
$W+U_f=U_i$
$W=508J$